What is the extreme y=x^4-16x^2 ?

The extreme y=x^4-16x^2 is Minimum(-2sqrt(2),-64),Maximum(0,0),Minimum(2sqrt(2),-64)

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extreme y=x^4-16x^2

extreme points

y = x^{4} - 16 x^{2}

Minimum (- 22, - 64), Maximum (0, 0), Minimum (22, - 64)

Solution steps

f^{'} (x) = 4 x^{3} - 32 x

Find intervals:

Decreasing

: - \infty < x < - 22,

Increasing

: - 22 < x < 0,

Decreasing

: 0 < x < 22,

Increasing

: 22 < x < \infty

Plug

x = - 22

into

x^{4} - 16 x^{2} : - 64

Minimum (- 22, - 64)

Plug

x = 0

into

x^{4} - 16 x^{2} : 0

Maximum (0, 0)

Plug

x = 22

into

x^{4} - 16 x^{2} : - 64

Minimum (22, - 64)

Minimum (- 22, - 64), Maximum (0, 0), Minimum (22, - 64)

What is the extreme y=x^4-16x^2 ?
The extreme y=x^4-16x^2 is Minimum(-2sqrt(2),-64),Maximum(0,0),Minimum(2sqrt(2),-64)